Sequence and Series – IB Math AA SL

Solution:

The n-th term of an arithmetic sequence is given by \( a_n = a_1 + (n-1)d \). Here, \( a_1 = 3 \) and \( d = 4 \). So:

\( a_5 = 3 + (5-1)\times 4 = 3 + 4 \times 4 = 3 + 16 = 19.\)

Answer: 19

Solution:

By definition of an arithmetic sequence, the second term is \( a_2 = a_1 + d \). Given \( a_1 = 2 \) and \( a_2 = 5 \), we have:

\( 5 = 2 + d \implies d = 3.\)

Answer: 3

Solution:

The sum of the first \( n \) terms of an arithmetic sequence is \( S_n = \frac{n}{2}(2a_1 + (n-1)d) \).

Here, \( n = 4 \), \( a_1 = 1 \), \( d = 2 \). Plugging in: \[ S_4 = \frac{4}{2} \bigl(2\times1 + (4-1)\times 2\bigr) = 2 \bigl(2 + 3\times2\bigr) = 2 \bigl(2 + 6\bigr) = 2 \times 8 = 16. \]

Answer: 16

Solution:

In a geometric sequence, the ratio between consecutive terms is constant. Here, \[ r = \frac{\text{second term}}{\text{first term}} = \frac{6}{3} = 2. \]

Answer: \( r = 2 \)

Solution:

The n-th term of a geometric sequence is \( a_n = a_1 r^{n-1} \).

For \( n=3 \), \( a_1 = 5 \), and \( r = \frac{1}{2} \), we get: \[ a_3 = 5\left(\frac{1}{2}\right)^{2} = 5 \times \frac{1}{4} = \frac{5}{4}. \]

Answer: \(\frac{5}{4}\)

Solution:

First, identify \( a_1 \) and \( r \). Here, \( a_1 = 2 \) and \( r = \frac{6}{2} = 3 \).

The sum of the first \( n \) terms of a geometric series is \[ S_n = a_1 \frac{1 - r^n}{1 - r}. \]

For \( n = 3 \): \[ S_3 = 2 \times \frac{1 - 3^3}{1 - 3} = 2 \times \frac{1 - 27}{-2} = 2 \times \frac{-26}{-2} = 2 \times 13 = 26. \]

Answer: 26

Solution:

Each term is obtained by multiplying the previous term by \(-\frac{1}{2}\).

• \( a_1 = 8 \)
• \( a_2 = 8 \times \left(-\frac{1}{2}\right) = -4 \)
• \( a_3 = -4 \times \left(-\frac{1}{2}\right) = 2 \)
• \( a_4 = 2 \times \left(-\frac{1}{2}\right) = -1 \)

Answer: \( 8, -4, 2, -1 \)

Solution:

Substitute \( n = 4 \) into the formula: \[ a_4 = 7 + (4-1)\times 5 = 7 + 3 \times 5 = 7 + 15 = 22. \]

Answer: 22

Solution:

The first two terms are \( a_1 \) and \( a_1 + 4 \). Their sum is given as 16:

\( a_1 + (a_1 + 4) = 16 \implies 2a_1 + 4 = 16 \implies 2a_1 = 12 \implies a_1 = 6.\)

Answer: 6

Solution:

The ratio \( r \) is obtained from consecutive terms: \[ r = \frac{\frac{1}{2}}{1} = \frac{1}{2}. \]

The n-th term is \( a_n = a_1 r^{n-1} \). Hence, \[ a_4 = 1 \times \left(\frac{1}{2}\right)^{3} = \frac{1}{8}. \]

Answer: \( r = \frac{1}{2}, \; a_4 = \frac{1}{8}. \)

Solution:

Let the first term be \( a_1 \) and the common difference be \( d \). We know:

• \( a_7 = a_1 + 6d = 43 \)
• \( a_{12} = a_1 + 11d = 68 \)

Subtract the first equation from the second: \[ (a_1 + 11d) - (a_1 + 6d) = 68 - 43 \] \[ 5d = 25 \implies d = 5. \]

Substitute \( d = 5 \) into \( a_7 = a_1 + 6d = 43 \): \[ a_1 + 6(5) = 43 \implies a_1 + 30 = 43 \implies a_1 = 13. \]

Answer: \( a_1 = 13 \) and \( d = 5 \).

Solution:

Use the sum formula: \[ S_5 = \frac{5}{2} \bigl(2a_1 + (5-1)d \bigr) = 40. \] Substituting \( d = 2 \): \[ \frac{5}{2}\left( 2a_1 + 4 \times 2 \right) = 40. \] \[ \frac{5}{2}\left( 2a_1 + 8 \right) = 40. \] \[ 5\left(2a_1 + 8\right) = 80. \] \[ 2a_1 + 8 = \frac{80}{5} = 16. \] \[ 2a_1 = 8 \implies a_1 = 4. \]

Answer: 4

Solution:

For a geometric sequence: \[ a_3 = a_1 r^{2} = 24, \quad a_6 = a_1 r^{5} = 192. \]

Dividing the second equation by the first: \[ \frac{a_1 r^{5}}{a_1 r^{2}} = \frac{192}{24} \implies r^{3} = 8 \implies r = 2. \]

Now substitute \( r = 2 \) into \( a_1 r^{2} = 24 \): \[ a_1 \times 2^2 = 24 \implies a_1 \times 4 = 24 \implies a_1 = 6. \]

Answer: \( a_1 = 6 \), \( r = 2 \).

Solution:

The nth term for a geometric sequence is \( a_n = a_1 r^{n-1} \). We have:

• \( a_2 = a_1 r = 9 \)
• \( a_5 = a_1 r^{4} = 729 \)

Divide the second by the first: \[ \frac{a_1 r^4}{a_1 r} = \frac{729}{9} \implies r^3 = 81 \implies r = 3 \quad (\text{assuming }r>0). \]

We would also consider negative roots if the problem allowed them, but typically if not stated, we take \( r = 3 \).

Answer: 3

Solution:

Here, \( a_1 = 2 \) and \( d = 3 \). The sum of the first \( n \) terms is: \[ S_n = \frac{n}{2}\left[ 2a_1 + (n-1)d \right]. \] For \( n = 10 \): \[ S_{10} = \frac{10}{2} \bigl[2 \times 2 + (10-1)\times 3\bigr]. \] \[ S_{10} = 5 \bigl[4 + 9 \times 3\bigr] = 5 \bigl[4 + 27\bigr] = 5 \times 31 = 155. \]

Answer: 155

Solution:

Use \( S_n = a_1 \frac{1-r^n}{1-r} \). For \( n = 5 \): \[ S_5 = 2 \times \frac{1 - (-3)^5}{1 - (-3)}. \] \[ (-3)^5 = -243, \] so \[ S_5 = 2 \times \frac{1 - (-243)}{1 + 3} = 2 \times \frac{1 + 243}{4} = 2 \times \frac{244}{4} = 2 \times 61 = 122. \]

Answer: 122

Solution:

The 7th term is: \[ a_7 = a_1 + 6d = -1 + 6d = 41 \implies 6d = 42 \implies d = 7. \]

Now use the sum formula: \[ S_7 = \frac{7}{2} [2 \times (-1) + (7-1)\times7]. \] \[ S_7 = \frac{7}{2} [-2 + 6 \times 7]. \] \[ S_7 = \frac{7}{2} [-2 + 42] = \frac{7}{2} [40] = 7 \times 20 = 140. \]

Answer: 140

Solution:

Let \( a_1 \) be the first term and \( d \) be the common difference. Then:

• \( a_4 = a_1 + 3d = 12 \)
• \( a_8 = a_1 + 7d = 24 \)

Subtract the first from the second: \[ (a_1 + 7d) - (a_1 + 3d) = 24 - 12 \implies 4d = 12 \implies d = 3. \]

Now substitute \( d = 3 \) into \( a_4 = a_1 + 3d = 12 \): \[ a_1 + 9 = 12 \implies a_1 = 3. \]

We need \( S_8 \). Using the sum formula: \[ S_8 = \frac{8}{2}[2a_1 + (8-1)d] = 4[2 \times 3 + 7 \times 3]. \] \[ S_8 = 4[6 + 21] = 4 \times 27 = 108. \]

Answer: 108

Solution:

The n-th term is \( a_n = a_1 r^{n-1} \). For \( n=5 \): \[ a_5 = -2 \left(\frac{3}{2}\right)^{4}. \] \[ \left(\frac{3}{2}\right)^4 = \frac{3^4}{2^4} = \frac{81}{16}. \] \[ a_5 = -2 \times \frac{81}{16} = -\frac{162}{16} = -\frac{81}{8}. \]

Answer: \(-\frac{81}{8}\)

Solution:

We see \( a_1 = 4 \) and \[ r = \frac{a_2}{a_1} = \frac{2}{4} = \frac{1}{2}. \]

Using the sum formula for a geometric series: \[ S_6 = a_1 \frac{1 - r^6}{1 - r} = 4 \times \frac{1 - \left(\frac{1}{2}\right)^6}{1 - \frac{1}{2}}. \] \[ \left(\frac{1}{2}\right)^6 = \frac{1}{64}. \] \[ S_6 = 4 \times \frac{1 - \frac{1}{64}}{\frac{1}{2}} = 4 \times \frac{\frac{63}{64}}{\frac{1}{2}} = 4 \times \frac{63}{64} \times 2 = 4 \times 2 \times \frac{63}{64} = 8 \times \frac{63}{64} = \frac{504}{64} = \frac{63}{8}. \]

Answer: \(\frac{63}{8}\)

Solution:

Here, \( a_1 = 5 \) and \( d = 5 \). The sum of the first \( n \) terms is: \[ S_n = \frac{n}{2}[2a_1 + (n-1)d]. \] Substitute \( a_1 = 5 \) and \( d = 5 \): \[ 1000 = \frac{n}{2} [2 \times 5 + (n-1) \times 5]. \] \[ 1000 = \frac{n}{2} [10 + 5n - 5] = \frac{n}{2} [5 + 5n]. \] \[ 1000 = \frac{n}{2} \times 5(n + 1). \] \[ 1000 = \frac{5n}{2} (n + 1). \] \[ 2000 = 5n (n + 1) = 5n^2 + 5n. \] \[ 5n^2 + 5n - 2000 = 0. \] \[ n^2 + n - 400 = 0. \] Solve this quadratic: \[ n = \frac{-1 \pm \sqrt{1 + 4 \times 400}}{2} = \frac{-1 \pm \sqrt{1601}}{2}. \] \(\sqrt{1601}\) is approximately 40.012..., so: \[ n = \frac{-1 \pm 40.012...}{2}. \] This gives two solutions: \[ n \approx \frac{-1 + 40.012}{2} \approx 19.506, \quad n \approx \frac{-1 - 40.012}{2} \approx -20.506. \] Since \( n \) must be an integer and positive (for the sum of the first \( n \) terms), we check if \( n = 19 \) or \( n = 20 \) might give us 1000. Let's check:

• For \( n = 19 \): \[ S_{19} = \frac{19}{2} [10 + 5 \times 18] = \frac{19}{2} [10 + 90] = \frac{19}{2} \times 100 = 950. \] Not 1000.
• For \( n = 20 \): \[ S_{20} = \frac{20}{2}[10 + 5 \times 19] = 10[10 + 95] = 10 \times 105 = 1050. \] Also not 1000.

Because the exact solution from the quadratic is about 19.506, not an integer, there are no integer solutions for \( n \). Therefore, there's no integer \( n \) that makes the sum exactly 1000.

Answer: No integer solutions.

Solution:

The sum of the first \( n \) terms of a geometric sequence is \[ S_n = a_1 \frac{1-r^n}{1-r}, \] for \( r \neq 1 \).

Here, \( a_1 = 10 \) and \( r = 2 \). Thus, \[ S_n = 10 \times \frac{1-2^n}{1-2} = 10 \times \frac{1-2^n}{-1} = -10 (1-2^n) = 10(2^n - 1). \] We want \[ S_n = 10(2^n - 1) \leq 120. \] \[ 2^n - 1 \leq 12. \] \[ 2^n \leq 13. \] We now find the largest integer \( n \) such that \( 2^n \leq 13 \):

• \(2^3 = 8 \leq 13\)
• \(2^4 = 16 \not\leq 13\)

Answer: \( n = 3 \).

Solution:

Here, \( a_1 = 6 \) and \( d = 5 \). The sum of the first \( n \) terms is: \[ S_n = \frac{n}{2} [2 \times 6 + (n-1)5]. \] We want the smallest \( n \) such that \( S_n > 500 \): \[ \frac{n}{2} [12 + 5(n-1)] > 500. \] \[ \frac{n}{2} [12 + 5n - 5] > 500. \] \[ \frac{n}{2} [7 + 5n] > 500. \] \[ \frac{n}{2} (5n + 7) > 500. \] \[ n (5n + 7) > 1000. \] \[ 5n^2 + 7n - 1000 > 0. \] Let's solve \( 5n^2 + 7n - 1000 = 0 \) approximately: \[ n = \frac{-7 \pm \sqrt{7^2 + 4 \times 5 \times 1000}}{2 \times 5} = \frac{-7 \pm \sqrt{49 + 20000}}{10} = \frac{-7 \pm \sqrt{20049}}{10}. \] \(\sqrt{20049} \approx 141.64\). So: \[ n = \frac{-7 + 141.64}{10} \approx 13.464 \quad\text{or}\quad n = \frac{-7 - 141.64}{10} \approx -14.864. \] Since \( n \) must be positive, we consider \( n \approx 13.464 \). For the inequality \( 5n^2 + 7n - 1000 > 0 \), the solution set will be \( n > 13.464 \) or \( n < -14.864 \).

We want the smallest integer \( n \) greater than 13.464, i.e., \( n = 14 \).

Let's check \( S_{13} \) and \( S_{14} \):

• \( S_{13} = \frac{13}{2}[12 + 5 \times 12] = \frac{13}{2}[12 + 60] = \frac{13}{2} \times 72 = 13 \times 36 = 468. \) This is not greater than 500.
• \( S_{14} = \frac{14}{2}[12 + 5 \times 13] = 7[12 + 65] = 7 \times 77 = 539. \) This exceeds 500.

Answer: 14 terms are needed.

Solution:

We see that the first term \( a_1 = 98 \) and \( r = \frac{1}{3} \). The sum of the first \( n \) terms is: \[ S_n = 98 \times \frac{1 - \left(\frac{1}{3}\right)^n}{1 - \frac{1}{3}} = 98 \times \frac{1 - \left(\frac{1}{3}\right)^n}{\frac{2}{3}} = 98 \times \frac{3}{2} \left[1 - \left(\frac{1}{3}\right)^n\right]. \] \[ S_n = 147 \left[1 - \left(\frac{1}{3}\right)^n\right]. \] We want \( S_n > 146 \): \[ 147 \left[1 - \left(\frac{1}{3}\right)^n\right] > 146. \] \[ 1 - \left(\frac{1}{3}\right)^n > \frac{146}{147}. \] \[ 1 - \left(\frac{1}{3}\right)^n > 0.993197\ldots \] \[ - \left(\frac{1}{3}\right)^n > 0.993197\ldots - 1 = -0.006802\ldots \] \[ \left(\frac{1}{3}\right)^n < 0.006802\ldots \] Take natural logs (or estimate powers of \(\frac{1}{3}\)): \[ \left(\frac{1}{3}\right)^5 = \frac{1}{243} \approx 0.004115\dots < 0.006802, \quad \left(\frac{1}{3}\right)^4 = \frac{1}{81} \approx 0.0123457\dots > 0.006802. \] So \( n = 5 \) is the smallest integer for which \( \left(\frac{1}{3}\right)^n < 0.006802\).

Therefore, \( n = 5 \) is the minimum integer such that \( S_n > 146 \).

Answer: \( n = 5 \).

Solution:

For a geometric series to have a finite sum to infinity, \(|r| < 1\). The sum to infinity is: \[ S_{\infty} = \frac{a_1}{1 - r} = 15. \] Substituting \( a_1 = 6 \): \[ \frac{6}{1 - r} = 15 \implies 6 = 15(1 - r) \implies 6 = 15 - 15r. \] \[ 15r = 15 - 6 = 9 \implies r = \frac{9}{15} = \frac{3}{5}. \]

Answer: \( r = \frac{3}{5} \).

Solution:

Let the first term be \( a_1 \) and the common difference be \( d \). Then:

• \( a_{10} = a_1 + 9d = 73 \)
• \( a_{20} = a_1 + 19d = 123 \)

Subtract the first from the second: \[ (a_1 + 19d) - (a_1 + 9d) = 123 - 73 \implies 10d = 50 \implies d = 5. \]

Substitute \( d = 5 \) back into \( a_{10} = a_1 + 9d = 73 \): \[ a_1 + 45 = 73 \implies a_1 = 28. \]

Now we find \( S_{30} \): \[ S_{30} = \frac{30}{2}[2a_1 + (30 - 1)d] = 15[2 \times 28 + 29 \times 5]. \] \[ = 15[56 + 145] = 15 \times 201 = 3015. \]

Answer: 3015

Solution:

Let \( a_1 \) be the first term and \( r \) be the common ratio. Then:

• \( a_2 = a_1 r = 12 \)
• \( a_6 = a_1 r^5 = \frac{3}{4} \)

From \( a_2 = 12 \), we have \( a_1 = \frac{12}{r} \). Substitute into \( a_6 = \frac{3}{4} \): \[ \left(\frac{12}{r}\right) r^5 = \frac{3}{4}. \] \[ 12 r^4 = \frac{3}{4}. \] \[ r^4 = \frac{3}{4 \times 12} = \frac{3}{48} = \frac{1}{16}. \] \[ r = \pm \frac{1}{2}. \]

For each \( r \), find \( a_1 \):

• If \( r = \frac{1}{2} \), then \( a_2 = a_1 \left(\frac{1}{2}\right) = 12 \implies a_1 = 24 \).
• If \( r = -\frac{1}{2} \), then \( a_2 = a_1 \left(-\frac{1}{2}\right) = 12 \implies a_1 = -24 \).

Answers:

• \( (a_1, r) = (24, \frac{1}{2}) \)
• \( (a_1, r) = (-24, -\frac{1}{2}) \)

Solution:

We have:

• \( a_1 = -4 \)
• \( a_4 = a_1 r^3 = 256 \)
• \( a_5 = a_1 r^4 = -1024 \)

From \( a_4 = -4 r^3 = 256 \): \[ r^3 = \frac{256}{-4} = -64. \] So \[ r = -4 \] (since \((-4)^3 = -64\)).

Let's verify with the 5th term: \[ a_5 = -4 \times (-4)^4 = -4 \times 256 = -1024. \] This matches perfectly.

Answer: \( r = -4 \).

Solution:

Let \( a_1 \) be the first term and \( d \) be the common difference. Then:

• \( a_5 = a_1 + 4d \)
• \( a_9 = a_1 + 8d \)

We are told \( a_5 = \frac{1}{3} a_9 \). Hence: \[ a_1 + 4d = \frac{1}{3} (a_1 + 8d). \] \[ 3(a_1 + 4d) = a_1 + 8d. \] \[ 3a_1 + 12d = a_1 + 8d. \] \[ 2a_1 + 4d = 0 \implies a_1 + 2d = 0 \implies a_1 = -2d. \]

Next, use the sum of the first 9 terms: \[ S_9 = \frac{9}{2}[2a_1 + (9-1)d] = 126. \] \[ S_9 = \frac{9}{2}[2a_1 + 8d] = 126. \] Substitute \( a_1 = -2d \): \[ 2a_1 + 8d = 2(-2d) + 8d = -4d + 8d = 4d. \] So \[ S_9 = \frac{9}{2} (4d) = 126. \] \[ 9 \times 2d = 126 \implies 18d = 126 \implies d = 7. \]

Then \( a_1 = -2d = -2 \times 7 = -14 \).

Answer: \( a_1 = -14, \; d = 7. \)

Solution:

Let the first term be \( a_1 = a \) and the common ratio be \( r \). We have:

• \( a_1 + a_2 = a + ar = 6 \) ... (1)
• \( a_1 + a_2 + a_3 + a_4 = a + ar + ar^2 + ar^3 = 90 \) ... (2)

From (1): \[ a(1 + r) = 6. \] We can express \( a = \frac{6}{1+r} \), provided \( r \neq -1 \).

Substitute into (2): \[ a + ar + ar^2 + ar^3 = a(1 + r + r^2 + r^3) = 90. \] \[ \frac{6}{1+r} [1 + r + r^2 + r^3] = 90. \] \[ 6 [1 + r + r^2 + r^3] = 90(1 + r). \] \[ 6 + 6r + 6r^2 + 6r^3 = 90 + 90r. \] \[ 6r^3 + 6r^2 + 6r + 6 = 90r + 90. \] \[ 6r^3 + 6r^2 + 6r + 6 - 90r - 90 = 0. \] \[ 6r^3 + 6r^2 - 84r - 84 = 0. \] Factor out 6: \[ 6(r^3 + r^2 - 14r - 14) = 0. \] \[ r^3 + r^2 - 14r - 14 = 0. \] We look for integer roots among factors of 14: ±1, ±2, ±7, ±14. By trial:

• \( r = 1 \) gives \( 1 + 1 - 14 - 14 = -26 \ne 0 \).
• \( r = 2 \) gives \( 8 + 4 - 28 - 14 = -30 \ne 0 \).
• \( r = -1 \) we must exclude because \( a = \frac{6}{1+r} \) would be undefined, but let's see if it even solves the polynomial: \(-1 + 1 + 14 - 14 = 0\). Actually, it does solve the polynomial, but it invalidates the expression for \(a\). So not a valid solution in the context of (1).
• \( r = 7 \) gives \( 343 + 49 - 98 - 14 = 280 \ne 0 \).
• \( r = -2 \) gives \( -8 + 4 + 28 - 14 = 10 \ne 0 \).
• \( r = -7 \) gives \( -343 + 49 + 98 - 14 = -210 \ne 0 \).

Since a simple integer guess is not working, let's attempt factoring by grouping or consider rational roots. Another approach: rewrite (1) and (2) in a different style.

Alternatively, we can write (2) with partial sums: \[ a_1 + a_2 + a_3 + a_4 = a(1 + r + r^2 + r^3). \] We know \( a_1 + a_2 = 6 \). Note that \( a_3 + a_4 = ar^2 + ar^3 = ar^2(1 + r) \).

So \[ (a_1 + a_2) + (a_3 + a_4) = 6 + ar^2(1 + r) = 90. \] \[ ar^2(1 + r) = 84. \] From (1), \( a(1 + r) = 6 \implies a = \frac{6}{1+r} \). So: \[ \frac{6}{1+r} r^2 (1 + r) = 84. \] \[ 6 r^2 = 84 \implies r^2 = 14. \] \[ r = \pm \sqrt{14}. \]

Now let's check each possibility with (1). If \( r = \sqrt{14} \), from (1): \[ a(1 + \sqrt{14}) = 6 \implies a = \frac{6}{1 + \sqrt{14}}. \] That is valid. If \( r = -\sqrt{14} \), then \[ a(1 - \sqrt{14}) = 6 \implies a = \frac{6}{1 - \sqrt{14}}. \] Both are mathematically possible solutions. The question does not restrict \( r \) to be positive, so both are valid.

Therefore, the solutions are: \[ a_1 = \frac{6}{1 + \sqrt{14}}, \quad r = \sqrt{14} \] or \[ a_1 = \frac{6}{1 - \sqrt{14}}, \quad r = -\sqrt{14}. \]

You can rationalize these denominators if needed, but usually leaving the result in fractional form with a surd is acceptable in an IB exam context.